Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct mv {
  int t, i, sz;
};
int n, m, a[110][110], sum, sum2, menor_i[110], menor_j[110];
vector<mv> ans, ans2;
void print(int quant, int idx) {
  vector<mv> v;
  if (idx == 1)
    v = ans;
  else
    v = ans2;
  printf("%d\n", quant);
  for (mv x : v) {
    for (int j = 0; j < x.sz; j++) {
      if (x.t == 0)
        printf("row ");
      else
        printf("col ");
      printf("%d\n", x.i + 1);
    }
  }
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) menor_i[i] = 9999999;
  for (int j = 0; j < m; j++) menor_j[j] = 9999999;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      scanf("%d", &a[i][j]);
      sum += a[i][j];
      menor_i[i] = min(menor_i[i], a[i][j]);
      menor_j[j] = min(menor_j[j], a[i][j]);
    }
  }
  sum2 = sum;
  int mv1 = 0, mv2 = 0;
  for (int i = 0; i < n; i++) {
    mv1 += menor_i[i];
    sum -= (menor_i[i] * m);
    ans.push_back(mv{0, i, menor_i[i]});
    for (int j = 0; j < m; j++) {
      menor_j[j] = min(menor_j[j], a[i][j] - menor_i[i]);
    }
  }
  for (int j = 0; j < m; j++) {
    mv1 += menor_j[j];
    sum -= (menor_j[j] * n);
    ans.push_back(mv{1, j, menor_j[j]});
  }
  for (int i = 0; i < n; i++) menor_i[i] = 9999999;
  for (int j = 0; j < m; j++) menor_j[j] = 9999999;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      menor_i[i] = min(menor_i[i], a[i][j]);
      menor_j[j] = min(menor_j[j], a[i][j]);
    }
  }
  for (int j = 0; j < m; j++) {
    mv2 += menor_j[j];
    sum2 -= (menor_j[j] * n);
    ans2.push_back(mv{1, j, menor_j[j]});
    for (int i = 0; i < n; i++) {
      menor_i[i] = min(menor_i[i], a[i][j] - menor_j[j]);
    }
  }
  for (int i = 0; i < n; i++) {
    mv2 += menor_i[i];
    sum2 -= (menor_i[i] * m);
    ans2.push_back(mv{0, i, menor_i[i]});
  }
  if (sum && sum2)
    puts("-1");
  else if (sum)
    print(mv2, 2);
  else if (sum2)
    print(mv1, 1);
  else if (mv1 > mv2)
    print(mv2, 2);
  else
    print(mv1, 1);
  return 0;
}